Question

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1) Independent tests are carried out to determine how far a two-man boat will travel on 4 liters of gas. A sample of 5 boats are tested and the 5 distances have a mean of 45.1 km and a standard deviation of 4.32 km. Construct a 99% confidence interval for the mean distances traveled for all two-man boats.

2) A survey is done and 512 homes are surveyed. 309 of them say they like cats.

a) Find p and q

b) Find the maximum error of estimate assuming a 98% confidence level.

Answer 1

Solution :

1) Given that,

Point estimate = sample mean = = 45.1

sample standard deviation = s = 4.32

sample size = n = 5

Degrees of freedom = df = n - 1 = 5 - 1 = 4

At 99% confidence level

= 1 - 99%

=1 - 0.99 =0.01

/2 = 0.005

t/2,df = t0.005,4 = 4.604

Margin of error = E = t/2,df * (s /n)

= 4.604 * ( 4.32 / 5)

Margin of error = E = 8.89

The 99% confidence interval estimate of the population mean is,

  ± E

= 45.1  ± 8.89

= ( 36.21, 53.99 )

2) Given that,

n = 512

x = 309

a) Point estimate = sample proportion = = x / n = 309 / 512 = 0.604

1 - = 1 - 0.604 = 0.396

b) At 98% confidence level

= 1 - 98%

=1 - 0.98 =0.02

/2 = 0.01

Z/2 = Z0.01  = 2.326

Maximum of error = E = Z / 2 * (( * (1 - )) / n)

= 2.326 (((0.604 * 0.396) / 512)

= 0.050